Waves

JEE Main level, numerical answer, medium

137 of 142

Question 137 of 142

The amplitude of wave disturbance propagating in the positive xx-direction is given by y=1(1+x)2y=\dfrac{1}{(1+x{)}^{2}} at time t=0t=0 and y=11+(x−2)2y=\dfrac{1}{1+(x-2{)}^{2}} at t=1st=1s, where xx and yy are in metres. The shape of wave does not change during the propagation. The velocity of the wave will be ms−1.m{s}^{-1}.

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Answer: 2

At t=0,y=11+x2t=0,y=\dfrac{1}{1+{x}^{2}}
At time t=t,y=11+(x−vt)2t=t,y=\dfrac{1}{1+(x-vt{)}^{2}}
At t=1,y=11+(x−v)2…(i)t=1,y=\dfrac{1}{1+(x-v{)}^{2}}\ldots (i)
At t=1,y=11+(x−2)2…(ii)t=1,y=\dfrac{1}{1+(x-2{)}^{2}}\ldots (ii)
Comparing (i)and(ii)(i)and(ii)
v=2ms−1v=2m{s}^{-1}

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